Let's say we have a curvilinear coordinate system $(\rho,\theta,\zeta)$. Also, let's say we have a surface $\Gamma$ parameterized as $\Gamma: \mathbf{S}(\rho(\theta,\zeta),\theta,\zeta) = \mathbf{S}(\theta,\zeta)$. I want to find the normal derivative $\partial_n$ of the normal unit vector $\mathbf{n}$ to the surface $\Gamma$. For this, let's use the following identity: $\nabla (\mathbf{A \cdot B}) = (\mathbf{A} \cdot \nabla) \mathbf{B} + (\mathbf{B} \cdot \nabla ) \mathbf{A} + \mathbf{A} \times( \nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})\\$. Let $\mathbf{A=B=n}$ to obtain the following:
$\nabla (\mathbf{n \cdot n}) = (\mathbf{n} \cdot \nabla) \mathbf{n} + (\mathbf{n} \cdot \nabla ) \mathbf{n} + \mathbf{n} \times( \nabla \times \mathbf{n}) + \mathbf{n} \times (\nabla \times \mathbf{n})$
$\rightarrow \quad 0 = 2\frac{\partial}{\partial n}\mathbf{n} + \mathbf{n} \times 0 + \mathbf{n} \times 0 $
$\rightarrow \quad \frac{\partial}{\partial n}\mathbf{n} = 0 $
The property $\nabla \times \mathbf{n}$ was applied to obtain the previous result. Now, in curvilinear coordinates, we have that $\partial_n = (\mathbf{n}\cdot \nabla) = (\mathbf{n}\cdot e^\rho)\partial_\rho + (\mathbf{n}\cdot e^\theta)\partial_\theta + (\mathbf{n}\cdot e^\zeta)\partial_\zeta$
Now, since the surface is parameterized as a function of $\theta,\zeta$, the normal vector at the surface of interest (which is our surface $\Gamma$) is simply defined as $\mathbf{n} = \frac{\partial_\theta \mathbf{S} \times \partial_\theta \mathbf{S}}{|\partial_\theta \mathbf{S} \times \partial_\theta \mathbf{S}|}$.
When evaluating $\partial_n \mathbf{n} = (\mathbf{n}\cdot e^\rho)\partial_\rho \mathbf{n} + (\mathbf{n}\cdot e^\theta)\partial_\theta \mathbf{n} + (\mathbf{n}\cdot e^\zeta)\partial_\zeta \mathbf{n}$
Should we completely ignore the term: $(\mathbf{n}\cdot e^\rho)\partial_\rho \mathbf{n} $ and set it to zero? In other words, is $(\mathbf{n}\cdot e^\rho)\partial_\rho \mathbf{n} = 0$ for any surface?
I have a surface parameterization and I am evaluating $(\mathbf{n}\cdot e^\theta)\partial_\theta \mathbf{n} + (\mathbf{n}\cdot e^\zeta)\partial_\zeta \mathbf{n}$ numerically but I am not getting exactly zero as the result. I was wondering if even though we know $\mathbf{n}$ on $\Gamma$, that does not mean that the vector $\mathbf{n}$ could not be a continuous vector in space that has derivatives in all directions?
I don't know if this makes sense. Any help would be appreciated. Thanks in advance!